Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The half-life of a radioactive isotope is three hours. If the initial mass of the isotope were \(256 \mathrm{~g}\), the mass of it remaining undecayed after 18 hours would be [2003] (a) \(4.0 \mathrm{~g}\) (b) \(8.0 \mathrm{~g}\) (c) \(12.0 \mathrm{~g}\) (d) \(16.0 \mathrm{~g}\)

Short Answer

Expert verified
The remaining mass of the isotope after 18 hours is 4.0 g.

Step by step solution

01

Understand the Concept of Half-Life

The half-life of a substance is the time it takes for half of the initial amount of the substance to decay. For this problem, the half-life is given as 3 hours, which means every 3 hours, half of the substance will have decayed.
02

Calculate Number of Half-Lives

To find out how many half-lives have passed in 18 hours, use the formula: \[ \text{Number of half-lives} = \frac{\text{Total time}}{\text{Half-life}} \] Substitute the given values: \[ \frac{18}{3} = 6 \] So, there are 6 half-lives in 18 hours.
03

Apply the Decay Formula Repeatedly

Starting from the initial mass of 256 g, halve the mass for each half-life elapsed. Keep halving the mass 6 times: 1. After 1 half-life, 256 g becomes 128 g. 2. After 2 half-lives, 128 g becomes 64 g. 3. After 3 half-lives, 64 g becomes 32 g. 4. After 4 half-lives, 32 g becomes 16 g. 5. After 5 half-lives, 16 g becomes 8 g. 6. After 6 half-lives, 8 g becomes 4 g.
04

Determine the Remaining Mass

After 18 hours, or 6 half-lives, the remaining mass of the isotope is 4 g.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

half-life
The half-life of a radioactive substance is a key concept in understanding how these materials decay over time. It represents the amount of time it takes for half of the initial quantity of a substance to decay. This means that after one half-life, only 50% of the original material remains. For instance, if you start with 100 grams of a radioactive substance, after one half-life you will have 50 grams left.
After two half-lives, this amount will halve again, leaving you with 25 grams. In the given exercise, the half-life is specified as 3 hours. This consistent rate of decay makes it easier to predict how much substance will be left after a given period. Understanding this concept helps in calculating the decay over longer periods using simple arithmetic.
isotope
An isotope is a variant of a chemical element that has the same number of protons but a different number of neutrons in the nucleus. This change in the number of neutrons does not affect the chemical properties but does affect the atomic mass and stability of the nucleus.
Radioactive isotopes are unstable as they have excess energy due to the imbalance between protons and neutrons. As a result, these isotopes try to reach a more stable state by emitting radiation in the form of particles or electromagnetic waves. Isotopes play a critical role in nuclear physics, medicine, and archaeology. For instance, carbon-14 is used in radiocarbon dating to determine the age of ancient artifacts. Each radioactive isotope has a unique half-life, which helps scientists understand how quickly they will decay.
decay formula
The decay formula helps in determining the remaining quantity of a radioactive substance after a certain period. It states that the remaining mass is computed by continuously halving the initial mass over successive half-lives. If you start with an initial mass, say 256 grams, you apply the formula:
- Remaining mass = Initial mass imes (1/2)^{( ext{Number of half-lives})} Where the number of half-lives is calculated by dividing the total elapsed time by the half-life. In our exercise, 18 hours corresponds to 6 half-lives (since 18 hours / 3 hours per half-life = 6).
Apply the decay formula: - 256 imes (1/2)^6 = 4 grams This straightforward formula simplifies the process of calculating radioactive decay, making it easy to determine the remaining quantity at any given time.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In one type of mutual annihilation of an electron and a positron, three \(\gamma\)-ray photons are produced. If each photon has an energy of \(0.3407 \mathrm{MeV}\), what is the mass of the positron in amu? (1 \(\mathrm{amu}=931.5 \mathrm{MeV}\) ) (a) \(7.986 \times 10^{-4}\) (b) \(5.486 \times 10^{-4}\) (c) \(16.86 \times 10^{-4}\) (d) \(2.243 \times 10^{-4}\)

If uranium (mass number 238 and atomic number 92 ) emits an \(\alpha\) particle, the product has mass number and atomic number (a) 234 and 90 (b) 236 and 92 (c) 238 and 90 (d) 236 and 90

A radioactive isotope has a half-life of 8 days. If today \(125 \mathrm{mg}\) is left over, what was its original weight 32 days earlier? (a) \(2 \mathrm{~g}\) (b) \(4 \mathrm{~g}\) (c) \(5 \mathrm{~g}\) (d) \(6 \mathrm{~g}\)

A heavy element has atomic number \(\mathrm{X}\) and mass number Y. Correct relation between \(\mathrm{X}\) and \(\mathrm{Y}\) is (a) \(\mathrm{X}=\mathrm{Y}\) (b) \(\mathrm{X}>\mathrm{Y}\) (c) \(\mathrm{X}<\mathrm{Y}\) (d) \(\mathrm{X}=Z(1-\mathrm{Y})\)

Match the following: List-I List-II (Reactions) (Particles) 1\. \({ }_{4} \mathrm{Be}^{9}+{ }_{2} \mathrm{He}^{4} \longrightarrow{ }_{6} \mathrm{C}^{12}+\ldots \ldots\) (i) \({ }_{2} \mathrm{He}^{4}\) 2\. \({ }_{6} \mathrm{C}^{12}+\ldots \ldots \longrightarrow{ }_{5} \mathrm{~B}^{10}+{ }_{2} \mathrm{He}^{4}\) (ii) \({ }_{0} \mathrm{n}^{1}\) \(3 .{ }_{7} \mathrm{~N}^{14}+\ldots \ldots \longrightarrow{ }_{8} \mathrm{O}^{17}+{ }_{1} \mathrm{H}^{1}\) (iii) \({ }_{1} \mathrm{D}^{2}\) 4\. \({ }_{20} \mathrm{Ca}^{40}+\ldots \ldots \longrightarrow{ }_{19} \mathrm{~K}^{37}+{ }_{2} \mathrm{He}^{4}\) (iv) \({ }_{1} \mathrm{H}^{1}\) The correct matching is: (a) (ii)(i) (iii) (iv) (b) (iii) (ii)(i) (iv) (c) (i) (ii) (iv) (iii) (d) (ii) (iii) (i) (iv)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free